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Suppose we build the hyperreal numbers $^*\Bbb R$ in the usual way as an ultrapower of the reals, built as a quotient of $\omega$-length sequences of reals mod some non-principal ultrafilter. Then, in general, we will get different hyperreal fields for different choices of ultrafilter.

However, there will be certain commonalities that all hyperreal fields share. For instance, no matter what ultrafilter you choose, we will always have that the sequence $\epsilon = (1/1, 1/2, 1/3, ...)$ is infinitesimal in the sense that we can show $\epsilon < 1/n$ for all $n \in \Bbb N$. Thus, we can also show that all hyperreal fields will have infinitesimal (and infinitely large) elements.

Thus, we may ask if there is some "least common denominator" of all the hyperreal fields. For instance: what is the largest field that injects into all of the hyperreals simultaneously? Clearly this will be strictly larger than the real numbers, since all hyperreals have infinitesimals. It will also have to be larger than the field of rational functions (or Laurent series?) $\Bbb R(\epsilon)$, since every hyperreal element has a square root, and thus so must $\epsilon$. The Puiseux series would seem to do, but is there anything strictly larger? The Levi-Civita field?

I am also interested in substructures larger than fields; in general if there is some "maximal substructure" with "as much transfer as possible" which is common to all of the hyperreals, although I may make a separate question for that.

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  • $\begingroup$ Well, I had suggested the field of Laurent series, but what about the field of rational functions in one infinitesimal $\epsilon$? That is a non-Archimedean field extension of the reals (the smallest?) which is the field of fractions in the polynomial ring $\Bbb R[\epsilon]$, and so as long as we are mapping $\Bbb R$ to $\Bbb R$, the entire thing is determined by where $\epsilon$ goes, isn't it? $\endgroup$ Commented Apr 26, 2023 at 3:42
  • $\begingroup$ Ah, yes, I was overcomplicating things. $\endgroup$ Commented Apr 26, 2023 at 3:43
  • $\begingroup$ (Still not sure if I understand well) If you ask the same question with $\mathbb C$ and require the embedding to be elementary, the answer is "yes": any extension of $\mathbb C$ with degree of transcendence 1 (they are all isomorphic) is minimal proper elementary extension. I guess the same holds for $\mathbb R$, or doesn't it? $\endgroup$ Commented Apr 26, 2023 at 11:08
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    $\begingroup$ Assuming CH, there is a unique hyperreal field up to isomorphism (since the ultrapower has cardinality $2^{\aleph_0} = \aleph_1$ and is $\aleph_1$-saturated and satisfies the complete theory of $\mathbb{R}$, and any two elementarily equivalent saturated structures of the same cardinality are isomorphic). So under CH your question has a trivial answer: the maximal field which embeds into all the hyperreal fields is the unique hyperreal field itself. So the answer to your question depends on the background set theory... $\endgroup$ Commented Apr 26, 2023 at 16:06
  • $\begingroup$ Alex Is right. Does rephrasing the question in terms of elementary extensions in place of ultrapowers comes closer to what you have in mind? $\endgroup$ Commented Apr 26, 2023 at 18:43

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Collecting some comments into an answer...

The answer to your question depends on the background set theory. If we assusme CH, then there is a unique hyperreal field up to isomorphism (since the ultrapower has cardinality $2^{\aleph_0} = \aleph_1$ and is $\aleph_1$-saturated and satisfies the complete theory of $\mathbb{R}$, and any two elementarily equivalent saturated structures of the same cardinality are isomorphic). So the maximal field which embeds into all the hyperreal fields is the unique hyperreal field itself.

You ask about languages other than the field language. To be precise, what the argument above proves is the following: assuming CH, if we view $\mathbb{R}$ as an $L$-structure in any countable language $L$, then for any two non-principal ultrafilters $U$ and $U'$ on $\omega$, the two ultrapowers $\mathbb{R}^\omega/U$ and $\mathbb{R}^\omega/U'$ are isomorphic as $L$-structures.

Without CH, one should be careful how to phrase the question. One way to interpret it is this: consider the embeddability pre-order on the class of all fields which embed into all hyperreal fields. It's not clear to me that this pre-order has a maximum class.

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